hdu--1312--Red and Black （dfs）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19953 Accepted Submission(s): 12122

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9

....#.

.....#

......

#@...#

.#..#.

11 9

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

7 7

..#.#..

###.###

...@...

###.###

..#.#..

0 0

Sample Output

 1 /*
 2     Name: hdu--1312--Red and Black 
 3     Copyright:  ©2017 日天大帝
 4     Author: 日天大帝
 5     Date: 30/04/17 19:19
 6     Description: dfs
 7 */
 8 #include<iostream>
 9 #include<cstring> 
10 using namespace std;
11 void dfs(int,int);
12 char map[30][30];
13 int w,h,ans;
14 int x,y;
15 int dir[4][2] = {0,1,0,-1,1,0,-1,0};
16 int main(){
17     ios::sync_with_stdio(false);
18 
19     while(cin>>w>>h,w||h){
20         memset(map,0,sizeof(map));
21         ans = 0;
22         for(int i=0; i<h; ++i){
23             cin>>map[i];
24             for(int j=0; j<w; ++j){
25                 if(map[i][j] == '@'){
26                     x = i;y = j;
27                 }
28             }
29         }
30         dfs(x,y);
31         cout<<ans<<endl;
32     }
33     return 0;
34 }
35 void dfs(int x,int y){
36     ans++;
37     for(int i=0; i<4; ++i){
38         int xx = x + dir[i][0];
39         int yy = y + dir[i][1];
40         if(xx<0 || yy<0 ||xx >=h|| yy>=w|| map[xx][yy] != '.')continue;
41         map[xx][yy] = '#';
42         dfs(xx,yy);
43     }
44 }